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2 years ago

Given the functions f(x) = 10x + 25 and g(x) = x + 8, which of the following functions represents f[g(x)] correctly?

Mathematics

1 answer:

Lapatulllka [165]2 years ago

6 0

Answer:

f[g(x)] = 10x + 105

Step-by-step explanation:

f(x) = 10x + 25 and g(x) = x + 8

$f[g(x)] = 10(x + 8) + 25 \\ \\ f[g(x)] = 10x + 80 + 25\\ \\ f[g(x)] = 10x + 105$

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PLEASE HELP QUICKLY! 20 points! I will mark best answer as Brainliest!

Alina [70]

t= 2.25 secs

Step-by-step explanation:

Step 1 :

Equation for motion with uniform acceleration is v = u+ at

where v is the final velocity

u is the initial velocity

a is the acceleration due to gravity

and t is the time

Step 2 :

Here , v = 0 because at the highest point final velocity is 0.

u = 72 feet/sec

a = -32 ft/sec^2

We need to find the time t.

Substituting in the equation we have,

0 = 72 -32 * t

=> 32 t = 72

=> t = 72/32 = 2.25 secs

4 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

0.249 in expanded form

nikklg [1K]

2.49×10^-1 is the answer

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3 years ago

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Scilla [17]

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3 years ago

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AfilCa [17]

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2 years ago